When AR(1) is expressed as MA($\infty$), I can interpret it as: let's say my wage this year depends only on last year's wage and a random shock (my boss' mood). But last year's wage also depends on the year before that, and so on. Therefore, my current wage is a accumulation of my boss' moods across years, with long-ago mood having smaller impact.

Is there a similar way to think about MA(1) as AR($\infty$)? Algebraically, I know that MA(1) can be expressed as AR($\infty$) like so. But I can't think of a way to interpret it.

$y_t = \epsilon_t + \phi \epsilon_{t-1}$

$y_t = \epsilon_t(1+\phi l)$

$(1+\phi l)^{-1}y_t = \epsilon_t$

$(1-\phi l + \phi^2l^2 - \phi^3l^3 + \cdots)y_t = \epsilon_t$

  • $\begingroup$ See Section 3.3 faculty.chicagobooth.edu/john.cochrane/research/papers/… $\endgroup$
    – John
    Apr 22, 2014 at 14:23
  • $\begingroup$ Thanks for the link, but I could not find an interpretation, only how to derive. $\endgroup$
    – Heisenberg
    Apr 22, 2014 at 14:26
  • $\begingroup$ Well I would interpret an MA as an MA, but if you have to interpret it as an AR, then just use the formula you derive. That last formula suggests that if you regress an MA(1) process as an AR(p), then you'd get specific regression coefficients. If $\phi$ is 0.1, then the first would be that, and then quickly decline to 0. Personally, I learned more about the difference between MA and AR by simulating random data and looking at PACFs. $\endgroup$
    – John
    Apr 22, 2014 at 14:46
  • $\begingroup$ I guess my confusion is: in MA(1), my $y_t$ only depends on the shock in this and last periods. Why in the world would it be related to $y_{t-q}$, which depends on shocks that are a long time ago? $\endgroup$
    – Heisenberg
    Apr 22, 2014 at 14:54
  • 1
    $\begingroup$ Keep in mind you're switching from thinking about $y_{t}$ as dependent on shocks to thinking about it being dependent on its past values. As an MA, $y_{t-1}$ depends on the $t-1$ shock and the $t-2$ shock. Hence, if you regress $y_{t}$ on $y_{t-1}$, then you're picking up part of the $t-1$ shock. However, you can't quite identify the whole shock because in that regression you only observe $y_{t-1}$, not the shocks. Adding more terms helps you identify the shocks. $\endgroup$
    – John
    Apr 22, 2014 at 15:08

1 Answer 1


MA(1): your salary depends on the random mood swings of your boss in this and previous period, he has a short memory. If you look at the salary this period, it is clearly correlated with a salary last period, because they share the same random mood swing in last period. However, the last period salary is also correlated with a salary two periods ago, since they share the random mood swing two periods ago. This in fact makes the current salary correlated with a salary two periods ago, and with salaries in distant past. That's why MA(1) is represented with AR($\infty$). Think of it as an infinite chain, where the link is the error term in period t which connects the observations in period t+1 and t.

  • 1
    $\begingroup$ superb explanation. $\endgroup$
    – forecaster
    Apr 22, 2014 at 15:57
  • $\begingroup$ So this is a case of correlation not causation between this period's salary and distant past's salary, yes? In contrast, the case of AR(1) as MA($\infty$) is actually causation, as in my boss mood in distant past does have a causal impact on my salary, albeit a diminished one. Plus, is there any intuition for the alternating sign? $\endgroup$
    – Heisenberg
    Apr 22, 2014 at 17:41
  • $\begingroup$ We don't normally associate random errors with causation. As to the alternative sign it simply comes from the Taylor expansion of $\frac{1}{1+\phi l}$ $\endgroup$
    – Aksakal
    Apr 22, 2014 at 17:54

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